HDU

Prairie dog comes again! Someday one little prairie dog Tim wants to visit one of his friends on the farmland, but he is as lazy as his friend (who required Tim to come to his place instead of going to Tim's), So he turn to you for help to point out how could him dig as less as he could. 

We know the farmland is divided into a grid, and some of the lattices form houses, where many little dogs live in. If the lattices connect to each other in any case, they belong to the same house. Then the little Tim start from his home located at (x0, y0) aim at his friend's home ( x1, y1 ). During the journey, he must walk through a lot of lattices, when in a house he can just walk through without digging, but he must dig some distance to reach another house. The farmland will be as big as 1000 * 1000, and the up left corner is labeled as ( 1, 1 ). 

Input

The input is divided into blocks. The first line in each block contains two integers: the length m of the farmland, the width n of the farmland (m, n ≤ 1000). The next lines contain m rows and each row have n letters, with 'X' stands for the lattices of house, and '.' stands for the empty land. The following two lines is the start and end places' coordinates, we guarantee that they are located at 'X'. There will be a blank line between every test case. The block where both two numbers in the first line are equal to zero denotes the end of the input.

Output

For each case you should just output a line which contains only one integer, which is the number of minimal lattices Tim must dig.

Sample Input

6 6..X...XXX.X.....X.X.....X.....X.X...3 56 30 0

Sample Output

3           

Hint

Hint: Three lattices Tim should dig: ( 2, 4 ), ( 3, 1 ), ( 6, 2 ).         

        题意：两只狗去约会，然后都比较懒，怎么办呢？谁追谁谁去呗 ，这不Prairie要去啦（）。

  还有这事怎么能公开呢，必须偷偷的进行。所以Prairie喜欢穿过房子，但是到了其他地方只好钻洞，不能让别人看见对不对~~~。挖洞要时间啊，那就加1.

但Prairie想更早的见到……。求最小时间。

这个呢·这里时间为上，所以要以时间优先排列。。。时间小得排前面(因为每个节点具有的权值不一样，普通的队列只是按简
单的步数，所以要用优先队列)

#include<stdio.h>#include<queue>#include<string.h>#include<algorithm>using namespace std;int n,m;char mapp[1100][1100];int vis[1100][1100];int sx,sy,ex,ey;struct note{    int x,y,s;    friend bool operator<(note a,note b)    {        return a.s>b.s;    // 注意：(   大于<   ) （  小于>  ）    }};int dir[4][2]= { {0,1},{-1,0},{0,-1},{1,0}};int bfs(){    priority_queue<note>Q;    note p,q;    p.x=sx;    p.y=sy;    p.s=0;    vis[sx][sy]=1;    Q.push(p);    while(!Q.empty())    {        p=Q.top();        Q.pop();        if(p.x==ex&&p.y==ey) return p.s;        for(int i=0; i<4; i++)        {            q=p;            q.x+=dir[i][0];            q.y+=dir[i][1];            if(q.x>=0&&q.x<n&&q.y>=0&&q.y<m&&!vis[q.x][q.y])            {                vis[q.x][q.y]=1;                if(mapp[q.x][q.y]=='.') q.s++;                Q.push(q);            }        }    }}int main(){    while(~scanf("%d%d",&n,&m)&&(n||m))    {        memset(vis,0,sizeof(vis));        for(int i=0; i<n; i++)            scanf("%s",mapp[i]);        scanf("%d%d%d%d",&sx,&sy,&ex,&ey);        sx--,sy--,ex--,ey--;        printf("%d\n",bfs());    }    return 0;}